Answer to Question #100862 in Molecular Physics | Thermodynamics for Pham Anh Khoa

Question #100862
An ideal gas (γ = 1.40) expands slowly and adiabatically. If the final temperature is one third the initial temperature, by what factor does the volume change?
1
Expert's answer
2020-01-03T09:23:24-0500

According to the adiabatic equation:


T1V1γ1=T2V2γ1T_1V_1^{\gamma-1} = T_2V_2^{\gamma-1}

T1T2=V2γ1V1γ1{T_1 \over T_2} = {V_2^{\gamma-1} \over V_1^{\gamma-1}}

If the final temperature was one third of the initial, then:


3=V2γ1V1γ13 = {V_2^{\gamma-1} \over V_1^{\gamma-1}}

If γ = 1.40:


3=V20.4V10.43 = {V_2^{0.4} \over V_1^{0.4}}

15.58850.4=V20.4V10.415.5885^{0.4} = {V_2^{0.4} \over V_1^{0.4}}

15.5885=V2V115.5885 = {V_2\over V_1}

That is, the gas expands 15.5885 times, compared with the initial volume.


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